 5v pin arduino due

Hi,

I am a novice in the arduino world.

I start my project with an arduino Leonardo but the SRAM was not enough to run the whole project. I am expecting to take a Arduino due which has much more SRAM.

The problem is that I was working with 5v LEDs on the Leonardo and the due is a 3v3 boards. Is it possible to run them trought the 5v pin of the due? or will it be fatal for the board ?

MC.

Hi,
Do the "5V LEDs" have a series resistor? Maybe no problem at all; just run from 3.3V

The problem is that I was working with 5v LEDs

There are no "5V" LEDs... With the same current limiting resistor, your LEDs will work at 3V, but they will be dimmer.

LEDs typically run at about 2V and they are "constant voltage", which means that under normal operating conditions, the voltage changes very little when the current changes*. like all diodes, they are non-linear. If the voltage is too low, very little current flows and they don't light up. If you exceed the normal operating voltage, excess current flows and they burn-out.

So, you control the current and the voltage falls into place. With regular-small LEDs you control the current with a resistor. Resistors are linear and per [u]Ohm's Law[/u], the current through a resistor is proportional to the voltage, and inversely proportional to the resistance.

The applied voltage is divided among series components (Kirchhoff's Laws). If we apply 5V and 2V is dropped across the LED, we know there is 3V across the resistor so we can calculate the current through the resistor. And, we know that the same current flow through all series components (Kirchhoff's Laws) so if we calculate the current through the resistor we also know the current through the LED.

Is it possible to run them trought the 5v pin of the due? or will it be fatal for the board ?

Yes, if the other end of the LED is connected to ground.

If you connect it to +5V and an I/O pin, the voltage across the LED (and series resistor) will switch between 5V and about 2V (the difference between the ~3.3V output and the 5V power supply) and it might not turn off completely, and it would be a "bad design". It might work, but the proper solution the right resistor.